Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 39}{x - 9} = \dfrac{3x + 15}{x - 9}$
Multiply both sides by $x - 9$ $ \dfrac{x^2 - 39}{x - 9} (x - 9) = \dfrac{3x + 15}{x - 9} (x - 9)$ $ x^2 - 39 = 3x + 15$ Subtract $3x + 15$ from both sides: $ x^2 - 39 - (3x + 15) = 3x + 15 - (3x + 15)$ $ x^2 - 39 - 3x - 15 = 0$ $ x^2 - 54 - 3x = 0$ Factor the expression: $ (x + 6)(x - 9) = 0$ Therefore $x = -6$ or $x = 9$ At $x = 9$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 9$, it is an extraneous solution.